I have the following code to display in Indian numbering system.
var x=125465778;
var res= x.toString().replace(/\B(?=(\d{3})+(?!\d))/g, ",");
Am getting this output :125,465,778.
I need output like this: 12,54,65,778.
Please help me to sort out this problem .
i'm late but i guess this will help :)
you can use Number.prototype.toLocaleString()
Syntax
numObj.toLocaleString([locales [, options]])
var number = 123456.789;
// India uses thousands/lakh/crore separators
document.getElementById('result').innerHTML = number.toLocaleString('en-IN');
// → 1,23,456.789
document.getElementById('result1').innerHTML = number.toLocaleString('en-IN', {
maximumFractionDigits: 2,
style: 'currency',
currency: 'INR'
});
// → ₹1,23,456.79
<div id="result"></div>
<div id="result1"></div>
For Integers:
var x=12345678;
x=x.toString();
var lastThree = x.substring(x.length-3);
var otherNumbers = x.substring(0,x.length-3);
if(otherNumbers != '')
lastThree = ',' + lastThree;
var res = otherNumbers.replace(/\B(?=(\d{2})+(?!\d))/g, ",") + lastThree;
alert(res);
Live Demo
For float:
var x=12345652457.557;
x=x.toString();
var afterPoint = '';
if(x.indexOf('.') > 0)
afterPoint = x.substring(x.indexOf('.'),x.length);
x = Math.floor(x);
x=x.toString();
var lastThree = x.substring(x.length-3);
var otherNumbers = x.substring(0,x.length-3);
if(otherNumbers != '')
lastThree = ',' + lastThree;
var res = otherNumbers.replace(/\B(?=(\d{2})+(?!\d))/g, ",") + lastThree + afterPoint;
alert(res);
Live Demo
Simple way to do,
1. Direct Method using LocalString()
(1000.03).toLocaleString()
(1000.03).toLocaleString('en-IN') # number followed by method
2. using Intl - Internationalization API
The Intl object is the namespace for the ECMAScript Internationalization API, which provides language sensitive string comparison, number formatting, and date and time formatting.
eg: Intl.NumberFormat('en-IN').format(1000)
3. Using Custom Function:
function numberWithCommas(x) {
return x.toString().split('.')[0].length > 3 ? x.toString().substring(0,x.toString().split('.')[0].length-3).replace(/\B(?=(\d{2})+(?!\d))/g, ",") + "," + x.toString().substring(x.toString().split('.')[0].length-3): x.toString();
}
console.log("0 in indian format", numberWithCommas(0));
console.log("10 in indian format", numberWithCommas(10));
console.log("1000.15 in indian format", numberWithCommas(1000.15));
console.log("15123.32 in indian format", numberWithCommas(15123.32));
if your input is 10000.5,
numberWithCommas(10000.5)
You will get output like this, 10,000.5
For integers only no additional manipulations needed.
This will match every digit from the end, having 1 or more double digits pattern after, and replace it with itself + ",":
"125465778".replace(/(\d)(?=(\d\d)+$)/g, "$1,");
-> "1,25,46,57,78"
But since we want to have 3 in the end, let's state this explicitly by adding extra "\d" before match end of input:
"125465778".replace(/(\d)(?=(\d\d)+\d$)/g, "$1,");
-> "12,54,65,778"
Given a number to below function, it returns formatted number in Indian format of digit grouping.
ex: input: 12345678567545.122343
output: 1,23,45,67,85,67,545.122343
function formatNumber(num) {
input = num;
var n1, n2;
num = num + '' || '';
// works for integer and floating as well
n1 = num.split('.');
n2 = n1[1] || null;
n1 = n1[0].replace(/(\d)(?=(\d\d)+\d$)/g, "$1,");
num = n2 ? n1 + '.' + n2 : n1;
console.log("Input:",input)
console.log("Output:",num)
return num;
}
formatNumber(prompt("Enter Number",1234567))
https://jsfiddle.net/scLtnug8/1/
I am little late in the game.
But here is the implicit way to do this.
var number = 3493423.34;
console.log(new Intl.NumberFormat('en-IN', { style: "currency", currency: "INR" }).format(number));
if you dont want currency symbol, use it like this
console.log(new Intl.NumberFormat('en-IN').format(number));
The easiest way is just to use Globalize plugin (read more about it here and here):
var value = 125465778;
var formattedValue = Globalize.format(value, 'n');
Try like below, I have found a number formatter Plugin here : Java script number Formatter
By using that i have done the below code, It works fine, Try this, It will help you..
SCRIPT :
<script src="format.20110630-1100.min.js" type="text/javascript"></script>
<script>
var FullData = format( "#,##0.####", 125465778)
var n=FullData.split(",");
var part1 ="";
for(i=0;i<n.length-1;i++)
part1 +=n[i];
var part2 = n[n.length-1]
alert(format( "#0,#0.####", part1) + "," + part2);
</script>
Inputs :
1) 125465778
2) 1234567.89
Outputs :
1) 12,54,65,778
2) 12,34,567.89
Simply use https://osrec.github.io/currencyFormatter.js/
Then all you need is:
OSREC.CurrencyFormatter.format(2534234, { currency: 'INR' });
// Returns ₹ 25,34,234.00
These will format the value in the respective systems.
$(this).replace(/\B(?=(?:\d{3})+(?!\d))/g, ',');
For US number system (millions & billions)
$(this).replace(/\B(?=(?:(\d\d)+(\d)(?!\d))+(?!\d))/g, ',');
For Indian number system (lakhs & crores)
This function can handle float value properly just addition to another answer
function convertNumber(num) {
var n1, n2;
num = num + '' || '';
n1 = num.split('.');
n2 = n1[1] || null;
n1 = n1[0].replace(/(\d)(?=(\d\d)+\d$)/g, "$1,");
num = n2 ? n1 + '.' + n2 : n1;
n1 = num.split('.');
n2 = (n1[1]) || null;
if (n2 !== null) {
if (n2.length <= 1) {
n2 = n2 + '0';
} else {
n2 = n2.substring(0, 2);
}
}
num = n2 ? n1[0] + '.' + n2 : n1[0];
return num;
}
this function will convert all function to float as it is
function formatAndConvertToFloatFormat(num) {
var n1, n2;
num = num + '' || '';
n1 = num.split('.');
if (n1[1] != null){
if (n1[1] <= 9) {
n2 = n1[1]+'0';
} else {
n2 = n1[1]
}
} else {
n2 = '00';
}
n1 = n1[0].replace(/(\d)(?=(\d\d)+\d$)/g, "$1,");
return n1 + '.' + n2;
}
Improvised Slopen's approach above, Works for both int and floats.
function getIndianFormat(str) {
str = str.split(".");
return str[0].replace(/(\d)(?=(\d\d)+\d$)/g, "$1,") + (str[1] ? ("."+str[1]): "");
}
console.log(getIndianFormat("43983434")); //4,39,83,434
console.log(getIndianFormat("1432434.474")); //14,32,434.474
Based on Nidhinkumar's question i have checked the above answers and
while handling negative numbers the output won't be correct for eg: -300 it should display as -300 but the above answers will display it as -,300 which is not good so i have tried with the below code which works even during the negative cases.
var negative = input < 0;
var str = negative ? String(-input) : String(input);
var arr = [];
var i = str.indexOf('.');
if (i === -1) {
i = str.length;
} else {
for (var j = str.length - 1; j > i; j--) {
arr.push(str[j]);
}
arr.push('.');
}
i--;
for (var n = 0; i >= 0; i--, n++) {
if (n > 2 && (n % 2 === 1)) {
arr.push(',');
}
arr.push(str[i]);
}
if (negative) {
arr.push('-');
}
return arr.reverse().join('');
Improvising #slopen's answer with decimal support and test cases.
Usage: numberToIndianFormat(555555.12) === "5,55,555.12"
utils.ts
export function numberToIndianFormat(x: number): string {
if (isNaN(x)) {
return "NaN"
} else {
let string = x.toString();
let numbers = string.split(".");
numbers[0] = integerToIndianFormat(parseInt(numbers[0]))
return numbers.join(".");
}
}
function integerToIndianFormat(x: number): string {
if (isNaN(x)) {
return "NaN"
} else {
let integer = x.toString();
if (integer.length > 3) {
return integer.replace(/(\d)(?=(\d\d)+\d$)/g, "$1,");
} else {
return integer;
}
}
}
utils.spec.ts
describe('numberToIndianFormat', () => {
it('nan should output NaN', () => {
expect(numberToIndianFormat(Number.NaN)).toEqual("NaN")
});
describe('pure integer', () => {
it('should leave zero untouched', () => {
expect(numberToIndianFormat(0)).toEqual("0")
});
it('should leave simple numbers untouched', () => {
expect(numberToIndianFormat(10)).toEqual("10")
});
it('should add comma at thousand place', () => {
expect(numberToIndianFormat(5555)).toEqual("5,555")
});
it('should add comma at lakh place', () => {
expect(numberToIndianFormat(555555)).toEqual("5,55,555")
});
it('should add comma at crore place', () => {
expect(numberToIndianFormat(55555555)).toEqual("5,55,55,555")
});
});
describe('with fraction', () => {
it('should leave zero untouched', () => {
expect(numberToIndianFormat(0.12)).toEqual("0.12")
});
it('should leave simple numbers untouched', () => {
expect(numberToIndianFormat(10.12)).toEqual("10.12")
});
it('should add comma at thousand place', () => {
expect(numberToIndianFormat(5555.12)).toEqual("5,555.12")
});
it('should add comma at lakh place', () => {
expect(numberToIndianFormat(555555.12)).toEqual("5,55,555.12")
});
it('should add comma at crore place', () => {
expect(numberToIndianFormat(55555555.12)).toEqual("5,55,55,555.12")
});
});
})
Indian money format function
function indian_money_format(amt)
{
amt=amt.toString();
var lastThree = amt.substring(amt.length-3);
var otherNumbers = amt.substring(0,amt.length-3);
if(otherNumbers != '')
lastThree = ',' + lastThree;
var result = otherNumbers.replace(/\B(?=(\d{2})+(?!\d))/g, ",") + lastThree;
alert(result)
return result;
}
indian_money_format(prompt("Entry amount",123456))
Related
I need to do a code to verify whether or not the entered number is an Armstrong number, but my code does not work for every number.
Could anyone tell me what am I missing? Are there any other ways to do this?
Thank you!
let e, x, d = 0;
let b = prompt("Enter a number");
x=b;
while (x > 0) {
e = x % 10;
x = parseInt(x/10);
d = d + (e*e*e);
}
if (b==d)
alert("given number is an armstrong number");
else
alert("given number is not an armstrong number");
<!DOCTYPE HTML>
<html>
<head>
<title>Armstrong</title>
</head>
<body>
</body>
</html>
I think the way you compute the result is wrong. According to Wikipedia, an Armstrong number, also called narcissistic number, has the following property:
[An Armstrong number] is a number that is the sum of its own digits each raised to the power of the number of digits.
You can compute it like this:
var number = prompt("Enter a number");
var numberOfDigits = number.length;
var sum = 0;
for (i = 0; i < numberOfDigits; i++) {
sum += Math.pow(number.charAt(i), numberOfDigits);
}
if (sum == number) {
alert("The entered number is an Armstrong number.");
} else {
alert("The entered number is not an Armstrong number.");
}
You can try this method, very easy to understand.
const armstrongNumber = (num) => {
const toArray = num.toString().split('').map(Number);
const newNum = toArray.map(a => {return a**3}).reduce((a, b) => a + b);
if(newNum == num){
console.log('This is an armstrong number');
}else{
console.log('This is not an armstrong number');
}
}
armstrongNumber(370);
//This is an armstrong number
Here is the solution to check Armstrong number without using Math Object.
function armstrongNum(number) {
const numberArr = String(number).split('');
const power = numberArr.length;
let TotalSum = numberArr.reduce((acc, cur) => {
return acc + (function(cur,power){
let curNum = Number(cur);
let product = 1;
while(power > 0) {
product *= curNum;
power --;
}
return product;
}(cur,power))
}, 0)
if (TotalSum === number) {
return true
}
return false
}
armstrongNum(153);
Here is an example of functional code for Armstrong Numbers.
<script>
function execute() {
var num1 = document.getElementById("Number1").value;
var num2 = document.getElementById("Number2").value;
var num3 = document.getElementById("Number3").value;
var concatNum = num1 + num2 + num3;
var num1Sqrt = num1 * num1 * num1;
var num2Sqrt = num2 * num2 * num2;
var num3Sqrt = num3 * num3 * num3;
if (num1Sqrt + num2Sqrt + num3Sqrt == concatNum) {
Answer.value = "The three integers you entered are Armstrong numbers.";
}
if (num1Sqrt + num2Sqrt + num3Sqrt != concatNum) {
Answer.value = "The three integers you entered are not Armstrong numbers.";
}
}
</script>
You first set your variables to what is entered, you then concatenate the string with the three integers.
You then square the three integers and set each value to a variable. You then have a basic check for equality, if the three squared values add up to your concatenated string then you have an Armstrong number.
This is how i solved mine:
function armstrong(num) {
var digits = num.toString().split('')
var realDigits = num
var a = 0
for (let i = 0; i < digits.length; i++){
digits[i] = Math.pow(digits[i], digits.length)
a += digits[i]
}
if (a == realDigits) {
console.log("Number is armstrong")
} else if (a != realDigits) {
console.log("Number is not armstrong")
}
}
armstrong(371)
//feel free to pass any value here
You can copy/paste and run this code at https://www.typescriptlang.org/play/
I hope this helps someone.
This works as well..
function isArmstrong (n) {
const res = parseInt(n, 10) === String(n)
.split('')
.reduce((sum, n) => parseInt(sum, 10) + n ** 3, 0);
console.log(n, 'is', res, 'Armstrong number')
return res
}
isArmstrong(153)
Here is another way to solve it.
let c = 153
let sum = 0;
let d = c.toString().split('');
console.log(d)
for(let i = 0; i<d.length; i++) {
sum = sum + Math.pow(d[i], 3)
}
if(sum == c) {
console.log("armstrong")
}
else {
console.log("not a armstrong")
}
Correct way to find Armstrong
var e, x, d = 0, size;
var b = prompt("Enter a number");
b=parseInt(b);
x=b;
size = x.toString().length;
while (x > 0) {
e = x % 10;
d = d + Math.pow(e,size);
x = parseInt(x/10);
}
//This is I solved without function
let num = prompt();
let num1 = num;
let sum = 0;
while(num > 0){
rem = num % 10;
sum = sum + Math.pow(rem, num1.length);
num = parseInt (num /10);
}
if (sum == num1) console.log("Armstrong");
else console.log("not Armstrong");
number is an Armstrong number or not.
let inputvalue=371
let spiltValue=inputvalue.toString().split('')
let output=0
spiltValue.map((item)=>output+=item**spiltValue.length)
alert(`${inputvalue} ${inputvalue==output? "is armstrong number" : "is not armstrong number"}`);
In order to get a Narcissistic/Armstrong number, you need to take the
length of the number as n for taking the power for summing the
value.
Here's another solution that works with an input >= 3 digits
(With Math.pow() each character is added as a number to the power of the array size)
const isArmstrongNumber = (n) => {
//converting to an array of digits and getting the array length
const arr = [...`${n}`].map(Number);
let power = arr.length;
const newNum = arr
.map((a) => {
return Math.pow(parseInt(a), power);
})
.reduce((a, b) => a + b);
return newNum == n;
};
console.log("Is it Armstrong? " + isArmstrongNumber(370));
console.log("Is it Armstrong? " + isArmstrongNumber(123));
console.log("Is it Armstrong? " + isArmstrongNumber(1634));
You can try for This, May this help for you:
<!DOCTYPE HTML>
<html>
<head>
<title>Armstrong</title>
</head>
<body>
</body>
</html>
<script>
var z,e,x,d=0;
var b=prompt("Enter a number");
x=parseInt(b);
while(x>0) //Here is the mistake
{
e=x%10;
x=parseInt(x/10);
d=d+(e*e*e);
console.log("d: "+d+" e: "+e);
}
if(parseInt(b)==d){
alert("given no is amstrong number");
}
else {
alert("given no is not an amstrong number");
}
</script>
Today i have problem in delimiting number for negative and positive number. For example, i have a textbox to insert my number and the result after write the number. Suddenly the number is separated with comma delimiter like this either the number is positive or negative
eg : 1000 -> 1,000
-1000 -> -1,000
-1000.12 -> -1,000.12
-0.00001 -> -0.00001
How to achieve this using javascript, what i know is using onkeypress and onkeyup. Thank you very much :)
This is not the best solution, but you can implement this according to your need.
var msg = document.getElementById('myInputBox'),
numberPattern = /^[0-9]+$/;
msg.addEventListener('keyup', function(e) {
msg.value = getFormattedData(msg.value);
});
function checkNumeric(str) {
return str.replace(/\,/g, '');
}
Number.prototype.format = function() {
return this.toString().replace(/(\d)(?=(\d{3})+(?!\d))/g, "$1,");
};
function getFormattedData(num) {
var i = checkNumeric(num),
isNegative = checkNumeric(num) < 0,
val;
if (num.indexOf('.') > -1) {
return num;
}
else if (num[num.length - 1] === ',') {
return num;
}
else if(i.length < 3) {
return i;
}else {
val = Number(i.replace(/[^\d]+/g, ''));
}
return (isNegative ? '-' : '') + val.format();
}
<input type="text" id='myInputBox' value="" />
Given Problem:
Write a function called "sumDigits".
Given a number, "sumDigits" returns the sum of all its digits.
var output = sumDigits(1148);
console.log(output); // --> 14
If the number is negative, the first digit should count as negative.
var output = sumDigits(-316);
console.log(output); // --> 4
My code:
function sumDigits(num) {
return num.toString().split("").reduce(function(a, b){
return parseInt(a) + parseInt(b);
});
}
My code solves the problem for positive integers. Any hints on how should I go about solving the problem for negative integers? Please and thanks.
Edit: And what if the given number is 0? Is it acceptable to add an if statement to return 0 in such cases?
Check to see if the first character is a -. If so, b is your first numeral and should be negative:
function sumDigits(num) {
return num.toString().split("").reduce(function(a, b){
if (a == '-') {
return -parseInt(b);
} else {
return parseInt(a) + parseInt(b);
}
});
}
You could use String#match instead of String#split for a new array.
function sumDigits(num) {
return num.toString().match(/-?\d/g).reduce(function(a, b) {
return +a + +b;
});
}
console.log(sumDigits(1148)); // 14
console.log(sumDigits(-316)); // 4
Somebody who is looking for a solution without reduce functions etc. can take this approach.
function sumDigits(num) {
var val = 0, remainder = 0;
var offset = false;
if (num <0) {
offset = true;
num = num * -1;
}
while (num) {
remainder = num % 10;
val += remainder;
num = (num - remainder) / 10;
}
if (offset) {
val -= 2 * remainder;//If the number was negative, subtract last
//left digit twice
}
return val;
}
var output = sumDigits(-348);
console.log(output);
output = sumDigits(348);
console.log(output);
output = sumDigits(1);
console.log(output);
//Maybe this help: // consider if num is negative:
function sumDigits(num){
let negativeNum = false;
if(num < 0){
num = Math.abs(num);
negativeNum = true;
}
let sum = 0;
let stringNum = num.toString()
for (let i = 0; i < stringNum.length; i++){
sum += Number(stringNum[i]);
}
if(negativeNum){
return sum - (Number(stringNum[0]) * 2);
// stringNum[0] has the "-" sign so deduct twice since we added once
} else {
return sum;
}
}
I need to increment a value similar to this:
A001 becomes A002
A999 becomes B001
B001 becomes B002
etc
Z999 becomes A001
I can increment an integer like this:
var x = 5;
x++;
Yields x = 6
I can increment an character like this:
var str = 'A';
str = ((parseInt(str, 36)+1).toString(36)).replace(/0/g,'A').toUpperCase();
if (str =='1A') {
str = 'A';
}
Yields the next character in the alphabet.
This code seems to work, but I'm not sure it's the best way?
var str = 'Z999';
if (str == 'Z999') {
results = 'A001';
}
else {
var alpha = str.substring(0,1);
num = str.substring(1,4);
if (alpha != 'Z' && num == '999') {
alpha= ((parseInt(alpha, 36)+1).toString(36)).replace(/0/g,'A').toUpperCase();
}
num++;
var numstr = num + "";
while (numstr .length < 3) numstr = "0" + numstr ;
if (numstr == 1000) {
numstr = '001';
}
results = alpha + numstr;
}
results seems to give the correct answer. Yes?
You could use parseInt(input.match(/\d+$/), 10) to extract the number at the end of the string and input.match(/^[A-Za-z]/) to retreive the single character at the beginning.
Increment and pad the number accordingly, and increment the character if the number is over 999 by retrieving the character's character code and incrementing that.
String.fromCharCode(letter.charCodeAt(0) + 1);
Full code/example:
function incrementNumberInString(input) {
var number = parseInt(input.trim().match(/\d+$/), 10),
letter = input.trim().match(/^[A-Za-z]/)[0];
if (number >= 999) {
number = 1;
letter = String.fromCharCode(letter.charCodeAt(0) + 1);
letter = letter === '[' ? 'A': (letter === '{' ? 'a' : letter);
} else {
number++;
}
number = '000'.substring(0, '000'.length - number.toString().length) + number;
return letter + number.toString();
}
document.querySelector('pre').textContent =
'A001: ' + incrementNumberInString('A001')
+ '\nA999: ' + incrementNumberInString('A999')
+ '\nB001: ' + incrementNumberInString('B001')
+ '\nB044: ' + incrementNumberInString('B044')
+ '\nZ999: ' + incrementNumberInString('Z999');
<pre></pre>
Output:
A001: A002
A999: B001
B001: B002
B044: B045
D7777: E001
Try storing A-Z in an array , using String.prototype.replace() with RegExp /([A-Z])(\d+)/g to match uppercase characters , digit characters . Not certain what expected result is if "Z999" reached ?
var arr = "ABCDEFGHIJKLMNOPQRSTUVWXYZ".split("");
var spans = document.querySelectorAll("span");
function count(el) {
var data = el.innerHTML.replace(/([A-Z])(\d+)/g, function(match, text, n) {
var _text, _n;
if (Number(n) === 999) {
_text = arr[ arr.indexOf(text) + 1 ];
} else {
_text = text
};
// `"Z999"` condition ?
if (_text === undefined) {
return "<mark>" + text + n + "</mark>"
}
_n = Number(n) + 1 < 1000 ? Number(n) + 1 : "001";
if (n < 10) {
return _text + n.slice(0, 2) + _n
};
if (n < 100) {
return _text + n.slice(0, 1) + _n
} else {
return _text + _n
}
});
el.innerHTML = data
}
for (var i = 0; i < spans.length; i++) {
count(spans[i])
}
<span>A001</span>
<span>A999</span>
<span>B001</span>
<span>C999</span>
<span>D123</span>
<span>Z999</span>
I'm looking for a way to construct regular expressions to match numeric inputs specified by a given integer range, ie. if I pass in a range of 1,3-4 then a regex would be returned matching just 1, 3 and 4.
I wrote the following method to try and do this:
function generateRegex(values) {
if (values == "*") {
return new RegExp("^[0-9]+$");
} else {
return new RegExp("^[" + values + "]+$");
}
}
I'm having issues however as sometimes I need to match double digits, such as "8-16", and I also need to ensure that if I am passed a single digit value, such as "1", that the generated regex matches only 1, and not say 11.
I really would like this to remain a pretty small snippet of code, but am not sure enough about regexs to know how to do this. Would be massively grateful for any help!
EDIT: I realise I wasn't clear, with my original paragraph, so have edited it. I realise the regex's that I originally generated do not work at all
Regexes don't know anything about numbers, only digits. So [8-16] is invalid because you say match between 8 and 1 (instead of 1 and 8 e.g.) plus the digit 6.
If you want to match numbers, you have to consider them lexically. For example, to match numbers between 1 and 30, you have to write something like (other regexes exist):
/^(30|[1-2]\d|[1-9])$/
I was sure it was 4-8 hours :-) In the end (and in its uselessness) it was a good exercise in composing Regexes. You are free to try it. If we exclude one use of continue and the use of the Array constructor, it's fully jsLint ok.
var BuildRegex = function(matches) {
"use strict";
var splits = matches.split(','),
res = '^(',
i, subSplit, min, max, temp, tempMin;
if (splits.length === 0) {
return new RegExp('^()$');
}
for (i = 0; i < splits.length; i += 1) {
if (splits[i] === '*') {
return new RegExp('^([0-9]+)$');
}
subSplit = splits[i].split('-');
min = BuildRegex.Trim(subSplit[0], '0');
if (min === '') {
return null;
}
if (subSplit.length === 1) {
res += min;
res += '|';
continue;
} else if (subSplit.length > 2) {
return null;
}
max = BuildRegex.Trim(subSplit[1], '0');
if (max === '') {
return null;
}
if (min.length > max.length) {
return null;
}
// For 2-998 we first produce 2-9, then 10-99
temp = BuildRegex.DifferentLength(res, min, max);
tempMin = temp.min;
if (tempMin === null) {
return null;
}
res = temp.res;
// Then here 100-998
res = BuildRegex.SameLength(res, tempMin, max);
}
res = res.substr(0, res.length - 1);
res += ')$';
return new RegExp(res);
};
BuildRegex.Repeat = function(ch, n) {
"use strict";
return new Array(n + 1).join(ch);
};
BuildRegex.Trim = function(str, ch) {
"use strict";
var i = 0;
while (i < str.length && str[i] === ch) {
i += 1;
}
return str.substr(i);
};
BuildRegex.IsOnlyDigit = function(str, start, digit) {
"use strict";
var i;
for (i = start; i < str.length; i += 1) {
if (str[i] !== digit) {
return false;
}
}
return true;
};
BuildRegex.RangeDigit = function(min, max) {
"use strict";
if (min === max) {
return min;
}
return '[' + min + '-' + max + ']';
};
BuildRegex.DifferentLength = function(res, min, max) {
"use strict";
var tempMin = min,
i, tempMax;
for (i = min.length; i < max.length; i += 1) {
tempMax = BuildRegex.Repeat('9', i);
res = BuildRegex.SameLength(res, tempMin, tempMax);
tempMin = '1' + BuildRegex.Repeat('0', i);
}
if (tempMin > tempMax) {
return null;
}
return {
min: tempMin,
res: res
};
};
BuildRegex.SameLength = function(res, min, max) {
"use strict";
var commonPart;
// 100-100
if (min === max) {
res += min;
res += '|';
return res;
}
for (commonPart = 0; commonPart < min.length; commonPart += 1) {
if (min[commonPart] !== max[commonPart]) {
break;
}
}
res = BuildRegex.RecursivelyAddRange(res, min.substr(0, commonPart), min.substr(commonPart), max.substr(commonPart));
return res;
};
BuildRegex.RecursivelyAddRange = function(res, prefix, min, max) {
"use strict";
var only0Min, only9Max, i, middleMin, middleMax;
if (min.length === 1) {
res += prefix;
res += BuildRegex.RangeDigit(min[0], max[0]);
res += '|';
return res;
}
// Check if
only0Min = BuildRegex.IsOnlyDigit(min, 1, '0');
only9Max = BuildRegex.IsOnlyDigit(max, 1, '9');
if (only0Min && only9Max) {
res += prefix;
res += BuildRegex.RangeDigit(min[0], max[0]);
for (i = 1; i < min.length; i += 1) {
res += '[0-9]';
}
res += '|';
return res;
}
middleMin = min;
if (!only0Min) {
res = BuildRegex.RecursivelyAddRange(res, prefix + min[0], min.substr(1), BuildRegex.Repeat('9', min.length - 1));
if (min[0] !== '9') {
middleMin = String.fromCharCode(min.charCodeAt(0) + 1) + BuildRegex.Repeat('0', min.length - 1);
} else {
middleMin = null;
}
}
middleMax = max;
if (!only9Max) {
if (max[0] !== '0') {
middleMax = String.fromCharCode(max.charCodeAt(0) - 1) + BuildRegex.Repeat('9', max.length - 1);
} else {
middleMax = null;
}
}
if (middleMin !== null && middleMax !== null && middleMin[0] <= middleMax[0]) {
res = BuildRegex.RecursivelyAddRange(res, prefix + BuildRegex.RangeDigit(middleMin[0], middleMax[0]), middleMin.substr(1), middleMax.substr(1));
}
if (!only9Max) {
res = BuildRegex.RecursivelyAddRange(res, prefix + max[0], BuildRegex.Repeat('0', max.length - 1), max.substr(1));
}
return res;
};
// ----------------------------------------------------------
var printRegex = function(p) {
"use strict";
document.write(p + ': ' + BuildRegex(p) + '<br>');
};
printRegex('*');
printRegex('1');
printRegex('1,*');
printRegex('1,2,3,4');
printRegex('1,11-88');
printRegex('1,11-88,90-101');
printRegex('1-11111');
printRegex('75-11119');
Test here http://jsfiddle.net/dnqYV/
The C# version is here http://ideone.com/3aEt3E
I'm not sure there is a (sane) way to test integer ranges with RegExp. I believe you're fixated on RegExp, where there are much simpler (more flexible) approaches. Take a look at IntRangeTest().
var range = new IntRangeTest('0,10-20');
console.log(
"0,10-20",
range.test("") == false,
range.test("-5") == false,
range.test("0") == true,
range.test("5") == false,
range.test("11") == true,
range.test("123.23") == false
);
If you feel like it, you can easily add this to Number.prototype. You could also quite easily make this an extension to RegExp, if that's what you're worried about.
Ok so it seems that there are 4 main cases that I need to address:
Single digits, ie 1, would simply generate the regex /^1$/
Multiple digits, ie 12, would require the regex /^12&/
Single digit ranges, ie 3-6, would generate the regex /^[3-6]$/
And finally, multiple digit ranges work in a similar method to multiple digits but with a range, ie 11-14 would become /^1[1-4]$/. These would need to be split into multiple regexes if they span over multiple start digits, Ie 23-31 would become /^2[3-9]|3[0-1]$/
Therefore, all I need to do is identify each of these cases and create a compound regex using | like xanatos suggested. Ie, to match all of the above criteria would generate a regex like:
/^( 1 | 12 | [3-6] | 1[1-4] | 2[3-9]|3[0-1] )$/
Do other agree this seems like a decent way to progress?